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Dynamic System Response Dynamic System Response • Solution of Linear, Constant-Coefficient, Ordinary Differential Equations – Classical Operator Method – Laplace Transform Method • Laplace Transform Properties • 1st-Order Dynamic System Time and Frequency Response • 2nd-Order Dynamic System Time and Frequency Response Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 1 Laplace Transform Methods • A basic mathematical model used in many areas of engineering is the linear ordinary differential equation with constant coefficients: d nq d n-1q dq a o + a o + +a o + a q = n dt n n-1 dt n-1 L 1 dt 0 o d mq d m-1q dq b i + b i + +b i + b q m dt m m-1 dt m-1 L 1 dt 0 i • qo is the output (response) variable of the system • qi is the input (excitation) variable of the system • an and bm are the physical parameters of the system Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 2 • Straightforward analytical solutions are available no matter how high the order n of the equation. • Review of the classical operator method for solving linear differential equations with constant coefficients will be useful. When the input qi(t) is specified, the right hand side of the equation becomes a known function of time, f(t). • The classical operator method of solution is a three-step procedure: – Find the complimentary (homogeneous) solution qoc for the equation with f(t) = 0. – Find the particular solution qop with f(t) present. – Get the complete solution qo = qoc + qop and evaluate the constants of integration by applying known initial conditions. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 3 • Step 1 – To find qoc, rewrite the differential equation using the differential operator notation D = d/dt, treat the equation as if it were algebraic, and write the system characteristic equation as: n n-1 a nD + a n-1D +L+a1D + a 0 = 0 – Treat this equation as an algebraic equation in the unknown D and solve for the n roots (eigenvalues) s1, s2, ..., sn. Since root finding is a rapid computerized operation, we assume all the roots are available and now we state rules that allow one to immediately write down qoc: – Real, unrepeated root s : 1 s1t qoc = c1e – Real root s2 repeated m times: s2t s2t 2 s2t m s2t qoc = c0e + c1te + c2t e +L+cmt e Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 4 – When the a’s are real numbers, then any complex roots that might appear always come in pairs a ± ib: at qoc = ce sinabt + ff – For repeated root pairs a ± ib, a ± ib, and so forth, we have: at at qoc = c0e sinabt + f0f + c1te sinabt + f1f+L – The c’s and f’s are constants of integration whose numerical values cannot be found until the last step. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 5 • Step 2 – The particular solution qop takes into account the "forcing function" f(t) and methods for getting the particular solution depend on the form of f(t). – The method of undetermined coefficients provides a simple method of getting particular solutions for most f(t)'s of practical interest. – To check whether this approach will work, differentiate f(t) over and over. If repeated differentiation ultimately leads to zeros, or else to repetition of a finite number of different time functions, then the method will work. – The particular solution will then be a sum of terms made up of each different type of function found in f(t) and all its derivatives, each term multiplied by an unknown constant (undetermined coefficient). Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 6 – If f(t) or one of its derivatives contains a term identical to a term in qoc, the corresponding terms should be multiplied by t. – This particular solution is then substituted into the differential equation making it an identity. Gather like terms on each side, equate their coefficients, and obtain a set of simultaneous algebraic equations that can be solved for all the undetermined coefficients. • Step 3 – We now have qoc (with n unknown constants) and qop (with no unknown constants). The complete solution qo = qoc + qop. The initial conditions are then applied to find the n unknown constants. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 7 • Certain advanced analysis methods are most easily developed through the use of the Laplace Transform. • A transformation is a technique in which a function is transformed from dependence on one variable to dependence on another variable. Here we will transform relationships specified in the time domain into a new domain wherein the axioms of algebra can be applied rather than the axioms of differential or difference equations. • The transformations used are the Laplace transformation (differential equations) and the Z transformation (difference equations). • The Laplace transformation results in functions of the time variable t being transformed into functions of the frequency-related variable s. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 8 • The Z transformation is a direct outgrowth of the Laplace transformation and the use of a modulated train of impulses to represent a sampled function mathematically. • The Z transformation allows us to apply the frequency- domain analysis and design techniques of continuous control theory to discrete control systems. • One use of the Laplace Transform is as an alternative method for solving linear differential equations with constant coefficients. Although this method will not solve any equations that cannot be solved also by the classical operator method, it presents certain advantages: – Separate steps to find the complementary solution, particular solution, and constants of integration are not used. The complete solution, including initial conditions, is obtained at once. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 9 – There is never any question about which initial conditions are needed. In the classical operator method, the initial conditions are evaluated at t = 0+, a time just after the input is applied. For some kinds of systems and inputs, these initial conditions are not the same as those before the input is applied, so extra work is required to find them. The Laplace Transform method uses the conditions before the input is applied; these are generally physically known and are often zero, simplifying the work. – For inputs that cannot be described by a single formula for their entire course, but must be defined over segments of time, the classical operator method requires a piecewise solution with tedious matching of final conditions of one piece with initial conditions of the next. The Laplace Transform method handles such discontinuous inputs very neatly. – The Laplace Transform method allows the use of graphical techniques for predicting system performance without actually solving system differential equations. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 10 • All theorems and techniques of the Laplace Transform derive from the fundamental definition for the direct Laplace Transform F(s) of the time function f(t): ¥ L f(t) = Fasf = f(t)e-stdt t > 0 z0 s = a complex variable = s + iw • This integral cannot be evaluated for all f(t)'s, but when it can, it establishes a unique pair of functions, f(t) in the time domain and its companion F(s) in the s domain. Comprehensive tables of Laplace Transform pairs are available. Signals we can physically generate always have corresponding Laplace transforms. When we use the Laplace Transform to solve differential equations, we must transform entire equations, not just isolated f(t) functions, so several theorems are necessary for this. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 11 • Linearity Theorem: L a1f1atf + a 2f2atf = L a1f1atf + L a2f2 atf = a1F1asf+ a 2F2 asf • Differentiation Theorem: df LL O = sFasf - fa0f NMdt QP 2 Ld f O 2 df L = s Fasf - sfa0f- a0f NMdt2 QP dt n n-1 Ld f O n n-1 n-2 df d f L = s Fasf - s fa0f - s a0f- - a0f NMdt n QP dt L dt n-1 – f(0), (df/dt)(0), etc., are initial values of f(t) and its derivatives evaluated numerically at a time instant before the driving input is applied. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 12 • Integration Theorem: a-1f Fasf f a0f L fatfdt = + z s s n a- kf a- nf Fasf f a0f L f atf = n + å n-k+1 s k=1 s a- nf n a-0f where f atf = zLz fatfadtf and f atf = fatf – Again, the initial values of f(t) and its integrals are evaluated numerically at a time instant before the driving input is applied. Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 13 • Delay Theorem: – The Laplace Transform provides a theorem useful for the dynamic system element called dead time (transport lag) and for dealing efficiently with discontinuous inputs. u(t) = 1.0 Þ t > 0 = Þ < -as u(t) 0 t 0 L f(t - a)u(t - a) = e F(s) u(t - a) = 1.0 Þ t > a u(t - a) = 0 Þ t < a Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 14 • Final Value Theorem: – If we know Q0(s), q0(¥) can be found quickly without doing the complete inverse transform by use of the final value theorem. limt®¥ f(t) = lims®0 sF(s) – This is true if the system and input are such that the output approaches a constant value as t approaches ¥. • Initial Value Theorem: – This theorem is helpful for finding the value of f(t) just after the input has been applied, i.e., at t = 0+.
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